## Algebra and Trigonometry 10th Edition

$x^2+(y-4)^2=16$ Center: $(0,4)$ Radius: $r=2$
$x^2+y^2-8y=0$ Notice that $-8y=-2(4)y$ and that $4^2=16$ $x^2+y^2-8y+16-16=0$ $x^2+(y^2-8y+16)=16$ $x^2+(y-4)^2=16$ The equation of a circle in standard form: $(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius $(x-0)^2+(y-4)^2=4^2$ Center: $(0,4)$ Radius: $r=2$